2016 11Th Jan Igcse Paper 3H

[Kipella]

Am I going mad.
My daughter tells me that there was a question today on the above paper as follows.
Given a rectangle with a diameter of 12cm and shortest side x cm.
Find the least and greatest possible values of x.
Has she missed a piece of information because I cannot do this.
Anyone out there able to help me please?????

[fiction-factory]

Did the question say diagonal rather than diameter?

[Captain2]

Suggest 'perimeter' and not 'diameter'?

[jim360]

In either case it's not entirely clear how you'd go about answering the question without some more information, although too much information would presumably constrain x to a particular value rather than give it a range of values.

If it were perimeter, then x could be anywhere between (just above) 0 and 3 cm (=12/4), and if it were a diagonal/ diameter then I suppose that would allow x to go between (just above) 0 and 6 Sqrt[2] cm (the last coming from Pythagoras' Theorem in the case that the rectangle is a square).

So it might be an answerable question, although I would like to see the paper for myself to be sure I've understood the question correctly and provide a more detailed answer, rather than just apparently pulling numbers out of thin air.

[fiction-factory]

Yes, the exact wording here is crucial

[Kipella]

Sorry should be diagonal.
Perimeter would surely constrain it to an exact answer. Wish I could see the paper.

[jim360]

Well, leaving the exact question aside for the moment, we could still look at what would happen if it was a perimeter. Then you'd have:

2 * (shortest side) + 2 * (longest side) = 12 cm

by adding all side lengths together, and then dividing by two:

shortest side + longest side = 6 cm

because this is the only equation we have, and there are two unknown quantities (the two side lengths), then so long as the two sides are both positive and shorter than 6 cm then the perimeter can be 12 for a whole set of possible side lengths. That would mean that the shortest side can be as short as basically no length at all -- and also can be no bigger than 3 cm (half of 6), because if it were any bigger then it would actually be the longest side. Hence x would lie between 0 and 3 cm, rather than being fixed at any particular value.

The same sort of reasoning would apply to the diagonal, although this time we would have:

(shortest side)^2 + (longest side)^2 = 144 cm^2

The shortest side can again be no larger than the longest side -- the limiting case is when they are (basically) equal, ie when:

(shortest side)^2 + (longest side)^2 = 2*(shortest side)^2 = 144 cm^2

Dividing by two and taking the square root would give the maximum length above, Sqrt[72] (or 6 Sqrt[2]); while the smallest length is again 0 cm.

I'm still not totally convinced that this is all there is to the question -- wouldn't be totally surprised if, say, x had to be a whole number or something.

[Buenchico]

As Jim360 has indicated, if it's 'diagonal' then the answer is simply
0 < x < Sqrt 6

[jim360]

6 Sqrt[2], not Sqrt[6], Buenchico.

[Buenchico]

Somehow
Sqrt (72) = 6 Sqrt (2)
got completely mis-typed in my post!
(I think I was far was too concerned with why Alt 251 failed to produce a square root sign, as it should have done, than to worry about little things like accuracy!)

Thanks, Jim ;-)

[Old_Geezer]

0 and 8.48528137423857...

As explained by others above:

The shortest distance is zero and the hypotenuse is then the same length as the third side.

The longest distance is calculated using Pythagoras' Theorem and realising that the greatest length the shortest size can be is when it is equal to the length of the third side (and the angles are 45º) as any bigger and it isn't the shortest side any more. √ (12² /2) = √72